I really do not like, and mostly avoid, circuits where potentiometers are used "backwards" and behave mostly like rheostats. I like to take the signal

*from *the wiper of a potentiometer, not inject a signal

*into* the wiper. But that's just the way I learned to use potentiometers. Your Mileage May Differ (YMMD).

In the circuit

@Agriias posted (which must be just a fragment of the actual real circuit), the so-called tone control will "short out" the audio signal from the plate of the amplifier tube through C1 when the wiper is rotated to the "ground" end of the potentiometer. Depending on the impedance of the plate circuit this may or may not have much effect on the speaker output volume because the reactance of C1 is fairly high at audio frequencies. The volume control does the same thing, shorting out the plate circuit of the amplifier tube when the potentiometer wiper is rotated to the "ground" position, but I guess this is okay if you like doing things that way. And of course you can't drive a speaker with both of its terminals shorted together. Yeah, I know... picky, picky, picky... and just when things were going along so well. I suggest Googleing (is that a word?) "audio tone control circuits" to get some idea of the many different ways people think tone controls should be implemented. Unfortunately, you will not find anything there on digital signal processing (DSP), which is the modern way to do it, but all of the analog circuits involve capacitor and resistor networks in various interesting arrangements, some quite clever.

But getting back to the OP's original question:

Hi all,

I have a question on capacitors and signal chains. If I have a signal that goes through two capacitors in parallel of different values such as .01uf o and .005nf what would be the result? Result depends on what circuit the two capacitors connect to next.

I understand that the lower .005nf capacitor would act as the path of least resistance to the higher frequencies No it doesn't. It acts as the capacitor with the highest reactance at any frequency., but does it short out all frequencies above its cutoff leaving only the remaining frequencies below that to go through the other capacitor while blocking anything below it as well? Nothing like this happens. Each capacitor passes a signal (current) according to its reactance and the remaining circuit to which it is connected. There is no sharp "cutoff frequency" only a gradual increase or decrease in signal response.

Another way of stating it is does the frequency essentially split off in both directions with redundancy through both capacitors? It does split off in both directions (Kirchoff's Law: what goes into a node must come out of a node), dividing according to Ohms Law. I don't know what you mean by "redundancy through both capacitors". The current distribution is what it is.

Also another question: if they were equal in value they would add together and have a mutual lower frequency cut off correct? with the same frequencies going through both capacitors? Yes, the would add together whether they were equal in value or not, and the same frequencies would go through both capacitors. I don't know what you mean by "a mutual lower frequency cut off." The frequency response depends on the presence of other components in the circuit, and it is what it is for any given value of capacitance, whether a single capacitor or two or more capacitors in parallel ... or any combination of series- and parallel-connected capacitors... as long as the combination results in the same value of capacitance.

Thanks,

Agriias

You must discard your idea that capacitors are somehow "frequency selective" with upper or lower cutoff frequencies. They are frequency selective only in the sense that their capacitive reactance is an inverse function of frequency: higher frequencies yield lower reactance and lower frequencies yield higher reactance. But reactance is a continuous function of frequency. There is no particular frequency that would result in an abrupt "cutoff" in frequency response.

Consider the following circuit: a capacitor connected in series with a resistor with an AC excitation voltage applied to this series combination. Measure the signal across the resistor (using whatever means at your disposal) and you will find that the signal increases with increasing frequency and decreases with decreasing frequency. At no frequency (except DC, which is not a frequency) does the signal totally disappear. At sufficiently high frequencies the signal will essentially be the same as the excitation signal because the capacitive reactance becomes a vanishingly small value compared to the resistance, which of course does not vary with frequency. At sufficiently low frequencies, there is a vanishingly small signal because the capacitive reactance becomes very large compared to the fixed resistance. Nowhere in between does the signal across the resistor vanish because a cutoff frequency is reached.

Then consider this: you can get the same frequency response from this resistor-capacitor circuit with ANY value of capacitor as long as you make the product of resistance (in ohms) and capacitance (in farads) the same. So, for example, a 1 μF capacitor and a 1 MegΩ resistor will yield the same frequency response, measured across the resistor, as a 10 μF capacitor and a 100 kΩ resistor, or an 0.1 μF capacitor and a 10 MegΩ resistor. The particular values you might select would depend on the characteristics of the following circuitry. The impedance of a 1 μF capacitor in series with a 1 MegΩ resistor is much lower than the impedance of an 0.1 μF capacitor in series with a 10 MegΩ resistor. You can verify this using an

on-line calculator. Just set the value for L to zero and plug in values for R and C.

73 de AC8NS

Hop